I want to calculate the Lipschitz constant of a continuous and piecewise linear function $f:[0,1]^2\rightarrow R$, like this \begin{equation*} f(x_1,x_2)=\left\{ \begin{aligned} 2x_1+x_2, &\quad\text{if} \quad x_1+x_2\leq 1\\ x_1+1, &\quad\text{if} \quad x_1+x_2>1 \end{aligned} \right. \end{equation*} I guess it is equal to the greatest Lipschitz constant among all pieces. Is there any textbook that contain related theorem?

  • $\begingroup$ Yes, but also check the case in which one point is in each piece. $\endgroup$ May 20, 2020 at 22:29
  • $\begingroup$ @Ramita I don't know how to prove it. I'm looking for a textbook on this issue. $\endgroup$ May 20, 2020 at 22:33
  • $\begingroup$ There is no well known theorem but it is not difficult to prove either. For the above it is $\sqrt{5}$ with the Euclidean norm. $\endgroup$
    – copper.hat
    May 20, 2020 at 22:36
  • $\begingroup$ @copper.hat I find a theorem of the vector-valued form for this issue, threesquirrelsdotblog.com/2018/03/16/…, and I feel the proof not easy. I want to cite such results, but I can not find any textbook that contain this issue. And it is not proper for me to cite a website. $\endgroup$ May 21, 2020 at 10:37

1 Answer 1


The problem has lots of structure, so there are many ways of approaching it. Here is one way:

Note that $f(x_1,x_2) = \min (2 x_1+x_2,x_1+1)$.

To see that the $\min$ of Lipschitz functions is Lipschitz:

Suppose $f_1,...,f_m$ are Lipschitz with rank $L$, then $f_k(x)-f_k(y) \le L \|x-y\|$ for all $k,x,y$. Then $\min_i f_i(x)-f_k(y) \le L \|x-y\|$ and choosing $k$ such that $\min_j f_j(y) = f_k(y)$ we see that $\min_i f_i(x)-\min_j f_j(y) \le L \|x-y\|$. Swapping $x,y$ shows that $\min_k f_k$ is Lipschitz with rank $L$. (This result is true more generally, but the finite case contains the basic idea.)

Note that $x \mapsto 2x_1+x_2$ has Lipschitz rank $\sqrt{5}$ and $x \mapsto x_1+1$ has Lipschitz rank $1$, so the smallest $L$ that will work is $L= \max(1,\sqrt{5})$.


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